Calculating this line integral (Finding the intersection curve and which parametrization to choose). 
Let $C$ be part of the intersection curve of the Paraboloid $z=x^2+y^2$ with the plane $2x+2y-z+2=0$ that starts from point $(3,1,10)$ and ends in $(1,3,10)$. 
We define $f(x,y,z)=z-y^2-2x-1$. 
Calculate $\int_{C}f dl$.

My work: 
Finding $C$: 
from the plane equation: $z=2x+2y+2$. 
Substituting that into the paraboloid equation: 
$2x+2y+2=x^2+y^2 \Longrightarrow x^2-2x+y^2-2y=2 \Longrightarrow (x-1)^2+(y-1)^2=4$. 
I find this result of getting a circle very weird, because the plane isn't parallel to $z=0$ plane, so I can't see why I received a circle, I expected an ellipse or something. The only thing I can think about is that I received the "Shadow" of the ellipse on the $xy$ plane, but I would appreciate any help understanding what have happened here! 

Anyway, I also got stuck here on which parametrization should I choose, if it's $x=r\cos(t), y=r\sin(t),z=r^2$ 
OR $x=1+r\cos(t),y=1+r\sin(t),z=?$ Then if I substitute it in the circle's equation I can find $z$. But I'm not sure if I can do that since $C$ isn't all the circle, it's just part of it. 

I would appreciate any help, thanks in advance! 
Edit After the help from answers: 
If I define $\vec r(t)=(1+2cos(t), 1+2sin(t), 4cos(t)+4sin(t)+6)$ to be the vector that draws the circle. 
Then $\vec r'(t) = (-2sin(t), -2cos(t), 4cos(t)-4sin(t))$. 
Where $\frac{\pi}{2} \ge t \ge 0$.  
$f(x,y,z)=z-y^2-2x-1=2x+2y+2-y^2-2x-1=-y^2+2y+1 = 2 - (y-1)^2$ 
And so my integral:
$$
\begin{split}
\int_C f dl
 &= \int_0^{\pi/2}
         2\cos(2t)
         \sqrt{(2\sin(t))^2 + (2\cos(t))^2 + (4\cos(t)-4\sin(t))^2} dt \\
 &= \int_0^{\pi/2}
         2\cos(2t)
         \sqrt{8 + (16\cos(t)^2 - 16\sin(2t) + 16\sin(t)^2)} dt \\
 &= \int_0^{\pi/2} 2\cos(2t)\sqrt{24-16\sin(2t)} dt
\end{split}
$$
I'm having some difficult times deciding how to do this integral
 A: The intersection curve of paraboloid and the slant plane is indeed an ellipse but it is an ellipse in the plane $2x+2y-z+2=0$. When you take the projection of each of the points on the intersection curve in xy-plane, it is a circle. In other words, the $x$ and $y$ coordinates of the points on the intersection curve follows $(x-1)^2 + (y-1)^2 = 4$ and $z$ coordinate can be written from the equation of the plane $z = 2x + 2y + 2$.
For you to visualize this, I will present another way to look at it. Take cylinder $(x-1)^2 + (y-1)^2 = 4$ and paraboloid $z = x^2 + y^2$. Their intersection curve is an ellipse in the plane $z = 2x + 2y + 2$. You can draw this in a 3D tool and see it.
Coming to parametrization, it is easier to parametrize as you mentioned later,
$r(t) = (1 + 2 \cos t, 1 + 2 \sin t, 6 + 4 \cos t + 4 \sin t), a \leq t \leq b$
$f(x,y,z)=z-y^2-2x-1 = 2y + 1 - y^2 = 2 - (y-1)^2$
So $f(r(t)) = 2 - 4 \sin ^2 t = 2 \cos 2 t$
Now find the bounds of $t$ plugging in start and end points which is pretty straightforward in this case. Then evaluate $|r'(t)| \ $ and finally follows the line integral.
A: You have a circle (not a disc) of radius $2$ centered at $(1,1)$, which would suggest
$$
\begin{split}
x &= 1 + 2\cos t \\
y &= 1 + 2\sin t
\end{split}
$$
from where $z = 2x+2y+2 = 6 + 4\cos t + 4\sin t$.
The last question is about the range of $t$. You need the part that starts at $(3,1,10)$ so you need $3 = x = 1 + 2\cos t$ and $1 = y = 1 + 2\sin t$, which implies $\cos t = 1$ and $\sin t = 0$, which you can solve for $t = t_0$.
Similarly, plug in the other end and solve for $t = t_1$. Then you integrate over $t \in [t_0, t_1]$.
